In this question, the OP asks to find the number of divisors of $2^2\cdot 3^3\cdot 5^3\cdot 7^5$ which are of the form $4n+1,n\in N$. The top answer points out that the required divisor is of the form $$3^a\cdot 5^b\cdot 7^c$$ with $0\leq a\leq 3,0\leq b\leq 3,0\leq c\leq 5$ and $a+c$ being even. The answer therefore is, apparently, $(4 \cdot 4 \cdot 6)/2=48$.
But this is wrong according to my book: the correct answer is $47$. Obviously, one case has been overcounted, but which? As far as I know, the person who wrote the top answer employed a fairly standard approach and should have arrived at the correct answer.